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My question is related to this one which tells us that fundamental group $\pi_{1}(X,x_0)$ is abelian $\textit{iff}$ for every pair $\alpha$ and $\beta$ of paths from $x_0$ to $x_1$, we have the same group isomorphism.

let $[I,X]$ denote the set of homotopy classes of maps of $I$ into $X$, where $I=[0,1]$. If $X$ is path connected, we can easily prove that $[I,X]$ has only one element, which mean that every path in $X$ is homotopic to each other (even a loop at $x_0 \in X$ is homotopic to path $p:x_0 \rightarrow x_1$).

using this fact, $\hat{\alpha}([f]) = [\bar{\alpha}]*[f]*[\alpha] = [\bar{\beta}]*[f]*[\beta]=\hat{\beta}([f]) $, wehre $\alpha$ and $\beta$ denote paths in path connected space $X$ ($\alpha \simeq \beta \Rightarrow [\alpha]= [\beta] $ ). So, it seems that there is no need for the fundamental group $\pi_{1}(X,x_0)$ to be abelian.

Also, for the proof that $\pi_{1}(X,x_0)$ is abelian, if you take 2 loops at $x_0$, then $[f]*[g]=[g]*[f]$ by virtue of being two paths in path-connected space $X$.

Am I wrong here? Where am I wrong?

MUH
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  • You are confusing free homotopies with base-point preserving homotopies. – Angina Seng Dec 01 '18 at 06:48
  • I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $\alpha, \beta : x_0 \rightarrow x_1$, then $\alpha \simeq \beta $(since $X$ is path connected) which imply that $ [\alpha] = [\beta]$ and hence, $\hat{\alpha}= \hat{\beta}$ – MUH Dec 01 '18 at 08:29
  • @LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $\hat{\alpha},\hat{\beta}$, which are base-point maps at $x_1$ or $\textit{loops}$ at $x_1$), doesn't it imply $\hat{\alpha}=\hat{\beta}$ – MUH Dec 01 '18 at 08:40

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Your conclusion that if $\alpha, \beta : x_0 \rightarrow x_1$, then $\alpha \simeq_p \beta$(since $X$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.

$p:x_0 \rightarrow x_1$ and $q:y_0 \rightarrow y_1$ are any two paths in path-connected space $X$. Then $p \simeq e_{x_0}$ and $q \simeq e_{y_0} $ and since, $e_{x_0} \simeq e_{y_{0}} \Rightarrow p \simeq q $. But this doesn't mean that $p \simeq_p q$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $t \in I$, which clearly doesn't get satisfied in the proof of homotopy ($\textit{free homotopy}$).

LearningMath
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